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36 votes

Let $p, q, r$ and $s$ be four primitive statements. Consider the following arguments:

- $P: [(¬p\vee q) ∧ (r → s) ∧ (p \vee r)] → (¬s → q)$
- $Q: [(¬p ∧q) ∧ [q → (p → r)]] → ¬r$
- $R: [[(q ∧ r) → p] ∧ (¬q \vee p)] → r$
- $S: [p ∧ (p → r) ∧ (q \vee ¬ r)] → q$

Which of the above arguments are valid?

- $P$ and $Q$ only
- $P$ and $R$ only
- $P$ and $S$ only
- $P, Q, R$ and $S$

42 votes

Best answer

An argument form is **valid **if no matter what propositions are substituted for the propositional

variables in its premises, the conclusion is true if the premises are all true.

i.e. $\left( p_1 \wedge p_2 \wedge \ldots \wedge p_n \right ) \rightarrow q$ is a tautology.

$$\begin{array}{l|l}\hline \text{$P:\large{\neg\color{Red}\not {p}\vee q}$} & \text{$Q:\large{\neg{p}}$}\\ \text{$\quad \large{\neg\color{Red}\not {r}\vee s}$} & \text{$\large{\quad \quad\color{Red}\not {q}}$}\\ \text{$ \large{\quad \underline{\color{Red}\not {p}\vee \color{Red}\not {r}}}$}& \text{$\large \quad {\underline {\neg\color{Red}\not {q}\vee(\neg {p}\vee r)}}$}\\ \text{$\large\qquad\;\; s\vee q$ also, $\large\neg s\to q$ $\large\color{Green}\checkmark$} & \text{$\qquad\quad\large\ \neg {p}\vee r$}\\ & \text{$\color{Red} {\large \neq \neg r}$} \\\hline

\text{$R: \large{\neg(q\wedge r)\vee p\\ \neg{q}\vee p}$$\large{ \;\;=\underline{\neg{q}\vee \neg{r}\vee {p}\\ \;\neg {q}\vee p}\\ \qquad {p}\vee\neg{q}\vee \neg {r}}$} & \text{$S:\large{\color{Red}\not {p}\\\neg\color{Red}\not {p}\;\vee \color{Red}\not {r}\\ \underline{q \vee \neg\color{Red}\not {r}}}$}\\ & \text{$\large\qquad\;\; q$ $\qquad\large\color{Green}\checkmark$}\\ \text{$\color{Red} {\large \neq r}$}\\\hline \end{array}$$

Correct Answer: $C$

edited
Dec 18, 2015
by Tendua

can u explain how we cut the terms. i am not knowing this method. either explain or give me a link. it will be thnkful.

what i learned from this . whenwver there is a and , change the line. in the same line only or terms are allowed.

cut down everything that is in complemented from also. and then just rite the common remaining terms. ,.

any other thing to keep in mind.

what i learned from this . whenwver there is a and , change the line. in the same line only or terms are allowed.

cut down everything that is in complemented from also. and then just rite the common remaining terms. ,.

any other thing to keep in mind.

0

@ Billy,It uses resolution principle-

If you have disjunction of literals given in the premises you can cut out the literals which are in true as well as complemented form in two different premises.

example- the various premises are:

pvqvr

pv¬qv¬r

--------------------

q and r can be cancelled and p is the conclusion.

Refer Keneth Rosen.

If you have disjunction of literals given in the premises you can cut out the literals which are in true as well as complemented form in two different premises.

example- the various premises are:

pvqvr

pv¬qv¬r

--------------------

q and r can be cancelled and p is the conclusion.

Refer Keneth Rosen.

13

I don't think this approach is right because even though we get one valid conclusion by using the resolution principle, there can be other valid conclusions too. For example, in S option, r can also be a conclusion. But, if you go by only the resolution principle, you will get q and mark S as wrong option. Correct me if I'm wrong.

0

If you don't found it easy to solve using resolution. Then use boolean logic minimization and try to get 1(TRUE) for validation.

eg:- $P:[(\neg P \vee Q)\wedge (r\rightarrow S)\wedge (P\vee R)] \Rightarrow (\neg S \rightarrow Q)$

$\equiv \overline{(\bar P+Q)(\bar R+S)(P+R)}\;+(\bar{\bar S}+Q)$

$\equiv (P\bar Q)+(R\bar S)+(\bar P \bar Q)+(S+Q)$

$\equiv \bar Q(P+\bar P)+(R \bar S)+(S+Q)$

$\equiv \bar Q + R\bar S + S+Q$

$\equiv TRUE \;(1)$

eg:- $P:[(\neg P \vee Q)\wedge (r\rightarrow S)\wedge (P\vee R)] \Rightarrow (\neg S \rightarrow Q)$

$\equiv \overline{(\bar P+Q)(\bar R+S)(P+R)}\;+(\bar{\bar S}+Q)$

$\equiv (P\bar Q)+(R\bar S)+(\bar P \bar Q)+(S+Q)$

$\equiv \bar Q(P+\bar P)+(R \bar S)+(S+Q)$

$\equiv \bar Q + R\bar S + S+Q$

$\equiv TRUE \;(1)$

2

8

17 votes

P: [(¬p v q) ∧ (r → s) ∧ (p v r)] → (¬s → q)

This must be valid ( All options contains P) No need to evaluate.

R: [[(q ∧ r) → p] ∧ (¬q v p)] → r ->

This is invalid, Put q = False, P= True, R = False to derive "False" result !

S: [p ∧ (p → r] ∧ (q v ¬ r)] → q This is valid !

->

p

P-> r

-------

r

q v ¬ r

----------

q

So Answer ->

3) P and S only

7 votes

According to me, using the rules of inference like resolution principle to come up with valid conclusions is tricky since there can be multiple valid conclusions. Instead, using proof by contradiction would be the most accurate way to tackle this question. To prove that a given statement is valid (i.e. a tautology), assume that it is not a tautology. It can only be possible if the right side is false and left side is true. Using this assumption, try to put truth values of proposition variables so as to make the left side true. If it is impossible to make left side true, then the statement is invalid. Otherwise, it's valid.

Another way is to convert the statement into a combination of negation, conjunction and disjunction operators and then simplify the statement as much as possible so that the end result is no more minimisable. If the result is a T, then it is valid. Else, it is invalid.

One more way is to draw the truth table of the statement. If all the values are true, then the statement is valid. Otherwise, it is not.

The last two methods are time-taking. The first method is the best.

Another way is to convert the statement into a combination of negation, conjunction and disjunction operators and then simplify the statement as much as possible so that the end result is no more minimisable. If the result is a T, then it is valid. Else, it is invalid.

One more way is to draw the truth table of the statement. If all the values are true, then the statement is valid. Otherwise, it is not.

The last two methods are time-taking. The first method is the best.

0

2 votes

let p->q is conditional proposition here p and q are compound propositions itself

Arguments to be valid if all combinations have to be tautology ( like T->T, F->T, F->F ) and its invalid if it have fallacy ( T->F )

if we somehow get this fallacy (T->F) then an argument is invalid

for options P and S u don't get any such combinations for T->F so P and S are VALID.

for option Q : if we put p=F,q=T,r=T then we get T->F so it is INVALID

for option R : if we put p=F,q=F,r=F then we get T->F so it is INVALID

so Ans is C.

Arguments to be valid if all combinations have to be tautology ( like T->T, F->T, F->F ) and its invalid if it have fallacy ( T->F )

if we somehow get this fallacy (T->F) then an argument is invalid

for options P and S u don't get any such combinations for T->F so P and S are VALID.

for option Q : if we put p=F,q=T,r=T then we get T->F so it is INVALID

for option R : if we put p=F,q=F,r=F then we get T->F so it is INVALID

so Ans is C.